# How to find the potential of a mixed Iron/ Cer solution?

Given the following half-reactions: \begin{align} \ce{Ce^{4+} + e^{−} &-> Ce^{3+}} & E° &= \pu{1.72 V}\\ \ce{Fe^{3+} + e^{−} &-> Fe^{2+}} & E° &= \pu{0.771 V}\\ \end{align}
A solution is prepared by mixing $$\pu{3.0 mL}$$ of $$\pu{0.30 M}$$ $$\ce{Fe^{2+}}$$ with $$\pu{2.0 mL}$$ of $$\pu{0.12 M}$$ $$\ce{Ce^{4+}}$$.

What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)? Calculate $$\ce{[Ce^{4+}]}$$ in the solution.

My attempt at solving the problem (which was incorrect):
First to find the potential (relative to SHE): \begin{align} E&=E^\circ-0.05916\cdot\log(\ce{Fe^{2+}/Fe^{3+}})\\ 0.771&=E^\circ-0.05916\cdot\log((3\cdot0.3)/(3\cdot0.3 + 2\cdot0.12))\\ E^\circ&=\pu{0.7649 V}\\ \end{align} I don't think this is correct.

Then: I am not sure how to find $$E^\circ$$ cerium or how to find $$\ce{[Ce^{4+}]}$$.

First it helps to have an idea of what will happen. Since 1.72 V is a lot more than 0.771 V (a difference of 1V in chemistry goes a long way), the numbers are telling us that $\ce{Ce^{4+}}$ is a much better oxidizer than $\ce{Fe^{3+}}$; in other words, in the presence of enough $\ce{Ce^{4+}}$, almost all $\ce{Fe^{2+}}$ will be oxidized to $\ce{Fe^{3+}}$. Conversely, in the presence of enough $\ce{Fe^{2+}}$, almost all $\ce{Ce^{4+}}$ will be reduced to $\ce{Ce^{3+}}$.

In equilibrium, therefore, either you'll have almost no $\ce{Fe^{2+}}$ left, or almost no $\ce{Ce^{4+}}$ left: one of the reagents will have been almost entirely consumed. To tell which one, look at the number of moles of each ion that went into the reaction at the moment of mixing. Keep in mind that each unit of $\ce{Fe^{2+}}$ reacts with one unit of $\ce{Ce^{4+}}$.

So now you've figured out which is the limiting reagent (the one that gets consumed almost entirely). Let's suppose for the sake of argument it's the $\ce{Fe^{2+}}$. Because you know how many moles of $\ce{Fe^{2+}}$ went in, and because all of it reacted, you know how many moles of $\ce{Ce^{4+}}$ got turned into $\ce{Ce^{3+}}$, and you can find the concentration of $\ce{Ce^{4+}}$ and $\ce{Ce^{3+}}$ in the equilibrium mix. And you also know the concentration of $\ce{Fe^{3+}}$, because essentially all the $\ce{Fe^{2+}}$ became $\ce{Fe^{3+}}$. That's all under the assumption that $\ce{Fe^{2+}}$ is the limiting reagent.

If the limiting reagent is $\ce{Ce^{4+}}$ instead, you can use the same logic to find the concentration of $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$ and $\ce{Ce^{3+}}$ in the equilibrium mix.

So now you have 3 concentrations out of 4. Finding the 4th requires knowing the equilibrium constant, which you can calculate from the difference in potentials. (Write the combined full reaction based on the half-reactions.)

You'll find that the 4th concentration is very small, justifying the claim that one of the reagents was almost entirely consumed.

If you're with me so far, please update the question with these results, and if you still need help with finding the potential at that point, we can take it from there...